Can anyone show me know to calculate the area and volume of a segment (circle)? (see attached file).
You can find the volume with or without calc. Which do you wanna do?.
Volume indicates 3-D. I suppose you mean a cylindrical tank of radius 13/20 and length 21/10?.
Here's the calc method for the volume(shaded portion) of the cylindrical tank of radius 13/20 and length 21/10.
$\displaystyle \frac{21}{5}\int_{\frac{-13}{20}}^{\frac{9}{20}}\sqrt{(\frac{13}{20})^{2}-x^{2}}dx=2.5154$
You can do it without by using the formula for the middle ordinate.
You are given the heighth of the segment, 1/5.
$\displaystyle h=R(1-cos(\frac{\theta}{2}))$
$\displaystyle \frac{1}{5}=\frac{13}{20}(1-cos(\frac{\theta}{2}))$
Solving, we see $\displaystyle {\theta}=2cos^{-1}(\frac{9}{13})$
Now, sub that into the formula for the segment of a circle:
$\displaystyle \frac{1}{2}r^{2}({\theta}-sin{\theta})$
$\displaystyle \frac{1}{2}(\frac{13}{20})^{2}(2cos^{-1}(\frac{9}{13})-sin(2cos^{-1}(\frac{9}{13})))\approx{0.1295145}$
This is the area of the segment, you need the shaded portion.
$\displaystyle {\pi}(\frac{13}{20})^{2}-0.1295145=1.1978$
Multiply this by the length, 21/10.
You get 2.5154. Same as above
The area of the shaded segment of the circle is equal to the area of the whole circle minus the area od the minor, unshaded, segment.
Area of circular segment is area of circular sector minus area of inscribed isosceles triangle in the sector.
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The area of the minor sector, where the minor segment is a part of:
radius = 1.3/2 = 0.65m
Area of sector = (1/2)(central angle, in radians)(radius^2)
Area of inscribed isosceles triangle = (1/2)(radius)(radius)*sin(central angle)
In the inscribed isosceles triangle,
cos(theta/2) = (0.65 -0.2)/(0.65) = 0.692307692
theta = 2[arccos(0.692307692)] = 1.6122283 radians
So, area of minor segment, A1, =
A1 = (1/2)(1.6122283)(0.65^2) -(1/2)(0.65)(0.65)sin(1.6122283)
A1 = 0.12951452 sq.m.
Therefore, area of major, shaded segment, A, =
A = pi(0.65^2) - A1
A = 1.1978 sq.m. -----------------answer.
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Volume in question,V,
V = A(2.1) = 2.5154 cu.m. -------------answer.
Does this assume that 1 cu.m = 1000 litres? If so, I thought the volume of a container was dependent on the mass and density of the liquid in it, according to the equation m = d x v (mass = density x volume). If so, then a 1 cu.m container that is filled with water will hold 1000 L of water (density of water = 1 g/cm3). However, if the container is filled with petrol (density of petrol = 750 g/L), then a 1 cu.m container filled with petrol will hold 750 L of petrol? Is this correct?